Two Digit Multiplier in U T method 6671 x 386

I would recommend you use the Bridge or the Vitruvian man method for 3 digit multipliers to reduce the cognitive load.

UT Method is recommended for two digit multipliers. However it is still possible to do the UT method with the a 3 digit multiplier.

Here is the illustration of the Method

If you want to multiply 6671 × 386 here are the steps.

Step 1: Add the same number of zeros as the number of digits in the multiplier.

So the multiplier 386 has 3 digits. So we add three zeroes in front of 6671 to get 000671

0 0 0 6 6 7 1 x 386

Step 2: Sum of two pair products is a single digit of the answer.

Use your middle and forefinger to keep track of whether to calculate U or the T. Place each finger in the middle of two pairs.

Step 2.1

Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
Using 8 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U8T8 = 00 + 00).
Using 3 in 386, the pair product of 0 and 6 is 0 + 1 = 1 (U3T3 = 00 + 18).
Adding the three number together we get 0 + 0 + 1 = 1.

U6 T6
=   U8 T8
=          U3 T3
0    0     0    6     6     7    1 x 386 = 1 _ _ _ _ _ _

Step 2.2

Moving to the next set of digits

Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
Using 8 in 386, the pair product of 0 and 6 is 0 + 4 = 4 (U8T8 = 00 + 48).
Using 3 in 386, the pair product of 6 and 6 is 8 + 1 = 9 (U3T3 = 18 + 18).
Adding the three number together we get 0 + 4 + 9 = 13.

We attach 3 in 13 as the next digit. But 13 has two digits so we carry over the 1 in 13. So the last digit from the previous step which is 1 becomes 2 (got by carrying over 1+1).

=   U6 T6
=         U8 T8
=                 U3 T3
0    0     0    6     6     7    1 x 386 = 2 3 _ _ _ _ _

Step 2.3

Moving to the next set of digits

Using 6 in 386, the pair product of 0 and 6 is 0 + 3 = 3 (U6T6 = 00 + 36).
Using 8 in 386, the pair product of 6 and 6 is 8 + 4 = 12 (U8T8 = 48 + 48).
Using 3 in 386, the pair product of 6 and 7 is 8 + 2 = 10 (U3T3 = 18 + 21).
Adding the three number together we get 3 + 12 +10 = 25.

We attach 5 in 25 as the next digit. But 25 has two digits, so we carry over the 2 in 25. So the last digit from the previous step which is 3 becomes 5 (got by carrying over 3 + 2).

=          U6 T6
=                 U8 T8
=                        U3 T3
0    0     0    6     6     7    1 x 386 = 2 5 5 _ _ _ _

Step 2.4

Moving to the next set of digits

Using 6 in 386, the pair product of 6 and 6 is 6 + 3 = 9 (U6T6 = 36 + 36).
Using 8 in 386, the pair product of 6 and 7 is 8 + 5 = 13 (U8T8 = 48 + 56).
Using 3 in 386, the pair product of 7 and 1 is 1 + 0 = 1 (U3T3 = 21 + 03).
Adding the three number together we get 9 + 13 +1 = 23.

We attach 3 in 23 as the next digit. But 23 has two digits, so we carry over the 2 in 23. So the last digit from the previous step which is 5 becomes 7 (got by carrying over 5 + 2).

=                 U6 T6
=                       U8 T8
=                              U3 T3
0    0     0    6     6     7    1 x 386 = 2 5 7 3 _ _ _

Step 2.5

Moving to the next set of digits

Using 6 in 386, the pair product of 6 and 7 is 6 + 4 = 10 (U6T6 = 36 + 42).
Using 8 in 386, the pair product of 7 and 1 is 6 + 0 = 6 (U8T8 = 56 + 08).
Using 3 in 386, the pair product of 1 and _ is 3 + _ = 3 (U3T3 = 03 + –-).
Adding the three number together we get 10 + 6 + 3 = 19.

We attach 9 in 19 as the next digit. But 19 has two digits, so we carry over the 1 in 19. So the last digit from the previous step which is 3 becomes 4 (got by carrying over 3 + 1).

=                      U6 T6
=                            U8 T8
=                                  U3 T3
0    0     0    6     6     7    1           x     386 = 2 5 7 4 9 _ _

Step 2.6

Moving to the next set of digits

Using 6 in 386, the pair product of 7 and 1 is 2 + 0 = 2 (U6T6 = 42 + 06).
Using 8 in 386, the pair product of 1 and _ is 8 + _ = 8 (U8T8 = 08 + –-).
Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –-).
Adding the three number together we get 2 + 8 + _ = 10.

We attach 0 in 10 as the next digit. But 10 has two digits, so we carry over the 1 in 10. So the last digit from the previous step which is 49 becomes 50 (got by carrying over 49 + 1).

=                              U6 T6
=                                   U8 T8
=                                             U3 T3
0    0     0    6     6     7    1                x     386 = 2 5 7 5 0 0 _

Step 2.7

Moving to the next set of digits

Using 6 in 386, the pair product of 1 and _ is 6 + _ = 6 (U6T6 = 06 + –-).
Using 8 in 386, the pair product of _ and _ is _ + _ = _ (U8T8 = –– + –-).
Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –-).
Adding the three number together we get 6 + _ + _ = 6.

6 becomes the last digit of the answer. .

=                                  U6 T6
=                                            U8 T8
=                                                   U3 T3
0    0     0    6     6     7    1                   x     386 = 2 5 7 5 0 0 6

This is your final answer.

Hope this clarifies.